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A292422 Numbers of the form x = 2^k*p*q, p,q odd primes, such that sigma(x)/phi(x) = 4. 4
418, 3596, 3956, 5396, 8636, 41656, 56536, 393104, 2072608, 2316448, 6543008, 17434528, 135394432, 217023616, 1264918784, 1490909824, 2710540544, 11444858368, 17669583104, 34797058304, 37698861568, 70572901376, 132968907776, 226965472256, 233356030976, 552070776832, 596357220352, 601188468736 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
A subset of A068390 and A020492 (balanced numbers). Conjectured to be infinite by Broughan and Zhou.
From David A. Corneth, Sep 21 2019: (Start)
Exactly 130 terms are of the form 2^35 * p * q.
We have phi and sigma are multiplicative and sigma(2^k) / phi(2^k) = 4 - 2/2^k, and sigma(p)/phi(p) = 1 + 2 / (p-1).
So we need (4 - 2/2^k) * (1 + 2 / (p-1)) <= 4 which gives a lower bound on p depending on k; p > nextprime(4*2^k).
We can then, given k and p, solve for q. Without loss of generality, p < q. Then search over the primes and stop for that value of k when p > q.
This method may be refined using insights from the article and/or given some k, solve the system (1 + 2 / (p-1)) * (1 + 2 / (q - 1)) = (a*m) / (b*m) for p and q where a/b is in lowest terms, m > 0. (End)
Furthermore, p < 8*2^k - 2. - David A. Corneth, Sep 26 2019
LINKS
Kevin A. Broughan and Qizhi Zhou, On the Ratio of the Sum of Divisors and Euler's Totient Function II, Journal of Integer Sequences, vol 17 (2014) article 14.9.2.
EXAMPLE
418 = 2*11*19; sigma(418) = 720 = 4*phi(418).
PROG
(PARI) is(n) = my(f = factor(n)); #f~ == 3 && f[2, 2] == 1 && f[3, 2] == 1 && f[1, 1] == 2 && sigma(f) / eulerphi(f) == 4 \\ David A. Corneth, Sep 21 2019
CROSSREFS
Subsequence of A293391 (sigma(x)/phi(x) square).
Sequence in context: A324613 A202527 A224557 * A328497 A138062 A350781
KEYWORD
nonn
AUTHOR
Jud McCranie, Sep 16 2017
STATUS
approved

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Last modified July 17 10:03 EDT 2024. Contains 374375 sequences. (Running on oeis4.)